Question

A small sandbag is dropped from rest from a hovering hot-air balloon. (assume the positive direction is upward.) (a) after 1.5 s, what is the velocity of the sandbag?

Answer 1

The velocity of the sandbag after 1.5 seconds is -14.7 m/s, downward.

Step-by-step explanation:

To find the velocity of the sandbag after 1.5 seconds, we can use the equation v = vo + at, where v is the final velocity, vo is the initial velocity (which is 0 since the sandbag is dropped from rest), a is the acceleration (which is equal to the acceleration due to gravity, approximately -9.8 m/s²), and t is the time.

Plugging in the values, we get v = 0 + (-9.8 m/s²)(1.5 s) = -14.7 m/s.

Therefore, the velocity of the sandbag after 1.5 seconds is -14.7 m/s, downward.

Answer 2

We know v0 = 0, a = 9.8, t = 4.0. We need to solve for v

so,

we use the equation:

v = v0 + at

v = 0 + 9.8*4.0

v = 39.2 m/s

Now we just need to solve for d, so we use the equation:

d = v0t + 1/2*a*t^2

d = 0*4.0 + 1/2*9.8*4.0^2

d = 78.4 m